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Let $B_t$ be a standard Brownian motion such that the processes $X_n(t)=e^{-nt}1_{[0,T]}(t)$, and

$\mathcal{H}$=$\{$($h_t$): $h_t$ is adapted, $\mathbb{E}\int_{0}^{\infty}h_t^2dt<\infty\}.$

How can we prove that $X_n \in \mathcal{H} $ and $\int_{0}^{T}X_n(t) dB_t\to 0$ in $L^2, n\to \infty$?

Here I tried to use distance function $d(e^{-nt},0)\to 0$ as $n\to \infty.$ i.e. $d(e^{-nt},0)=e^{-nt}$ and using Ito isometry $$\mathbb{E}\bigg(\int_{0}^{T}e^{-nt}dB_t\bigg)^2=\int_{0}^{T}\mathbb{E}\bigg(e^{-nt}\bigg)^2dt=\int_{0}^{T}\mathbb{E}\bigg(e^{-2nt}\bigg)dt$$. But, I couldn't go further.

I would appreciate any help? Thanks.

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    What is the expectation value of a constant....? – 2017-02-23

1 Answers 1

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Ohh Thanks. I got it now. My problem was well-definedness and expectation of constant(i.e. $X_n(t)$ is non-random). Now it is well-defined because $ \int_{0}^{T} e^{-2nt}dt=0$ as n goes to infinity, and $\int_{0}^{T}\mathbb{E}\bigg(e^{-nt}\bigg)^2dt=\int_{0}^{T}e^{-2nt}dt=0<\infty$. So isometry is satisfied. Hence, $\mathbb{E}\bigg(\int_{0}^{T}e^{-nt}dB_t\bigg)^2=\int_{0}^{T}\mathbb{E}\bigg(e^{-nt}\bigg)^2dt=0.$ Am I right?